Electronic Journal of Differential Equations, Vol. 2014 (2014), No. 89, pp. 1–10.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu
HAMILTONIANS REPRESENTING EQUATIONS OF MOTION
WITH DAMPING DUE TO FRICTION
STEPHEN MONTGOMERY-SMITH
Abstract. Suppose that H(q, p) is a Hamiltonian on a manifold M , and
L̃(q, q̇), the Rayleigh dissipation function, satisfies the same hypotheses as a
Lagrangian on the manifold M . We provide a Hamiltonian framework that
gives the equation
∂H
∂H
∂ L̃
(q, p), ṗ = −
(q, p) −
(q, q̇)
∂p
∂q
∂ q̇
The method is to embed M into a larger framework where the motion drives
a wave equation on the negative half line, where the energy in the wave represents heat being carried away from the motion. We obtain a version of Nöther’s
Theorem that is valid for dissipative systems. We also show that this framework fits the widely held view of how Hamiltonian dynamics can lead to the
“arrow of time.”
q̇ =
1. Introduction
The purpose of this document is to provide a Hamiltonian framework which gives
rise to equations of motion that include friction or damping effects. The author has
found a number of other works [2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12], but none of them
seem to solve the problem in the manner presented here. The methods of this paper
allow for extremely general formulas for the damping term, but the method is also
artificial. Nevertheless we are able to extract various principles such as Nöther’s
Theorem.
2. Notation
We represent the position of a particle by q, which might be in a finite or infinite
dimensional manifold M . (Everything in this document is formal, and there is no
attempt at rigor.) As usual we have a Lagrangian L(q, q̇) which is strictly convex in
the second coordinate. The equations of motion are the solution to the variational
equation δS = 0, where the action S is
Z T1
S=
L(q, q̇) dt;
(2.1)
T0
2000 Mathematics Subject Classification. 70H25.
Key words and phrases. Hamiltonian; Lagrangian; Rayleigh dissipation function;
friction; Nöther’s Theorem.
c
2014
Texas State University - San Marcos.
Submitted January 23, 2014. Published April 2, 2014.
1
2
S. MONTGOMERY-SMITH
EJDE-2014/89
that is,
d ∂L ∂L
−
=0
dt ∂ q̇
∂q
As usual, we create the Hamiltonian
H(q, p) = p · q̇ − L(q, q̇)
(2.2)
(2.3)
where p satisfies the equation
p=
∂L
∂ q̇
(2.4)
It can then be shown that
∂H
(2.5)
∂p
∂L
∂H
=−
(2.6)
∂q
∂q
where in the latter formula one has to be aware that the first partial derivative is
keeping q̇ constant, and the second partial derivative is keeping p constant. The
equation of motion becomes
∂H
(2.7)
ṗ = −
∂q
The usual example is
L(q, q̇) = 21 q̇ · A · q̇ − V (q)
(2.8)
where A = A(q) is positive definite symmetric. Then
q̇ =
p = A · q̇
H(q, p) =
1
2
p·A
−1
(2.9)
· p + V (q)
and the equation of motion is
∂A
∂V
d
(A · q̇) − 12 q̇ ·
· q̇ +
=0
dt
∂q
∂q
The book [1] gives many more details.
(2.10)
(2.11)
3. Striving to understand friction
Here we describe a widely held view of how friction arises from reversible Hamiltonian dynamics. Our concrete equations of motion obtained in Sections 4 and 5
should conform to the framework described here.
Suppose the manifold M is a submanifold of M0 ×M1 , where where M0 represents
the macroscopic motion, and M1 represents the microscopic motion. The position
then splits as q = (q0 , q1 ), and the momentum splits as p = p0 + p1 in the obvious
manner. Because M1 represents microscopic motion, it makes sense that M1 should
have much higher dimension than M0 , indeed we will often suppose that M1 is
infinite dimensional. For example, maybe the microscopic motion is a large number
of high frequency harmonic oscillators.
We assume that the Hamiltonian splits as
H(q, p) = H0 (q0 , p0 ) + H1 (q0 , q1 , p1 )
(3.1)
Here H0 represents the macroscopic motion without friction. The second part, H1 ,
is the energy of the microscopic motion plus coupling between the microscopic and
macroscopic parts. We assume that all the interaction between the microscopic and
macroscopic parts is via position and not momentum.
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HAMILTONIANS REPRESENTING EQUATIONS OF MOTION
3
Because H1 does not depend upon p0 , from equation (2.5) we obtain
∂H0
∂p0
∂H1
q̇1 =
∂p1
q̇0 =
(3.2)
(3.3)
and the Lagrangian splits as
L(q, q̇) = L0 (q0 , q̇0 ) + L1 (q0 , q1 , q̇1 )
(3.4)
L0 (q0 , q̇0 ) = p0 · q̇0 − H0 (q0 , p0 )
(3.5)
L1 (q0 , q1 , q̇1 ) = p1 · q̇1 − H1 (q0 , q1 , p0 )
(3.6)
where
and equations (2.4) and (2.6) become
∂L0
∂ q̇0
∂L1
=
∂ q̇1
∂H0
=−
∂q
∂H1
=−
∂q
p0 =
(3.7)
p1
(3.8)
∂L0
∂q
∂L1
∂q
(3.9)
(3.10)
(The only mildly non-obvious statements are equation (3.9) which follows from
∂L0 /∂q0 = −∂H0 /∂q0 and ∂L0 /∂q1 = ∂H0 /∂q1 = 0, and equation (3.10) which
follows by subtraction.)
The equations of motion now become
ṗ0 = −
ṗ1 =
∂H0
∂H1
−
∂q0
∂q0
∂H1
−
∂q1
(3.11)
(3.12)
Now, if the microscopic motion starts at rest, then we hope that the second law of
thermodynamics should be obeyed, and as a function of time, H0 should be nonincreasing, and H1 should be non-decreasing. Thus the hope is that the motion
described by q1 somehow provides a memory of what has happened to q0 , so that
we end up with
∂H1
= Φ(q0 , q̇0 )
(3.13)
∂q0
for some function Φ. Then H0 satisfies
∂H0
∂H0
+ ṗ0 ·
∂q0
∂p0
= q̇0 · (−ṗ0 − Φ(q0 , q̇0 )) + ṗ0 · q̇0 = −q̇0 · Φ(q0 , q̇0 )
Ḣ0 = q̇0 ·
(3.14)
If Φ satisfies a property like
y · Φ(x, y) ≥ 0
then we that H0 is non-increasing in time.
(3.15)
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S. MONTGOMERY-SMITH
EJDE-2014/89
The case we can handle with our approach is to consider
Φ(x, y) =
∂ L̃
(x, y)
∂y
(3.16)
where y 7→ L̃(x, y) is a strictly convex. Note that property (3.15) holds if L̃(x, y)
attains its minimum at y = 0, because
Z 1
∂ L̃
∂ L̃
∂ ∂ L̃
y·
y·
(x, y) = y ·
(x, 0) +
(x, λy) dλ
∂y
∂y
∂y
0 ∂λ
(3.17)
Z 1
2
∂ L̃
(x, λy) · y dλ ≥ 0
=
y·
∂y 2
0
since the Hessian of L̃(x, ·) is positive definite.
The initial values of q1 and p1 should be very important, perhaps representing
that the parts in M1 are at rest, or are oscillating in such a manner that the
oscillations can only increase. This is because Hamiltonian equations are reversible
in time. But dissipative systems should only be solvable forwards in time. This
is sometimes known as the “arrow of time.” For example, we know that if a wine
glass is thrown to the ground, then it shatters. The reverse process, where the wine
glass comes back together, is possible in theory if the initial positions and momenta
of the molecules and atoms are given in a very precise manner (and we neglect the
effects of quantum physics). But finding this initial data should be very difficult.
4. Friction via the wave equation: proof of concept
We will show how to artificially create damping via the wave equation. As a
proof of concept, let us explain how to create a Hamiltonian such that
∂H0
∂p0
∂H0
ṗ0 = −
− B · q̇0
∂q0
q̇0 =
(4.1)
(4.2)
for some positive definite matrix B.
An example of an infinite dimensional Hamiltonian is the wave equation on
C 1 (R, M ), whose Hamiltonian is given by
Z ∞
−1
1
· p(s) + 12 q 0 (s) · B · q 0 (s) ds
(4.3)
2 p(s) · B
−∞
0
Here q denotes ∂q/∂s. Now the equation of motion coming from this Hamiltonian
is
B q̈ = Bq 00
(4.4)
which by d’Alembert’s principle has a general solution
q(s, t) = φ1 (s + t) + φ2 (s − t)
(4.5)
This is a good example of a Hamiltonian for which Poincaré’s recurrence theorem
does not apply. And furthermore, the equations of motion are a point of variation
of the Lagrangian, but are clearly not a minimum of the Lagrangian:
Z ∞
·
0
1
1 0
(4.6)
2 q̇(s) · B q̇(s) − 2 q (s) · B · q (s) ds
−∞
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HAMILTONIANS REPRESENTING EQUATIONS OF MOTION
5
The way we simulate friction is to consider the macroscopic motion as driving a
wave equation on the half line. Let M1 = C 1 ((−∞, 0], M0 ), that is, the set of
functions q : (−∞, 0] → M0 such that q 0 is bounded, q 0 (s) → q 0 (0) as s → 0− ,
where q 0 (0) denotes the left derivative of q(s) at s = 0, and q 0 (s) → 0 as s → −∞.
Then let
M = {(q0 , q1 ) ∈ M0 × M1 : q1 (0) = q0 }
(4.7)
with the Hamiltonian
Z
0
H(q, p) = H0 (q0 , p0 ) +
−∞
1
2
p1 (s) · B −1 · p1 (s) +
1
2
q10 (s) · B · q10 (s) ds
(4.8)
We see that the equation of motion is
q̇0 =
∂H0
∂p0
(4.9)
∂H0
− B · q10 (0)
∂q0
B q̈1 = Bq100
ṗ0 = −
(4.10)
(4.11)
For example, to obtain equations (4.9) and (4.11), see that for any infinitesimal
perturbation δq = (δq0 , δq1 ) of q = (q0 , q1 ), noting that δq0 = δq1 (0) and δq10 (s) → 0
as s → −∞, we have
Z 0
∂H0
∂H
= δq0 ·
δq ·
(q0 , p0 ) +
δq10 (s) · B · q10 (s) ds
∂q
∂q0
−∞
Z 0
h
i0
∂H0
= δq0 ·
(q0 , p0 ) + δq1 (s) · B · q10 (s)
−
δq1 (s) · B · q100 (s) ds
∂q0
−∞
−∞
Z 0
∂H0
(q0 , p0 ) + δq0 · B · q10 (0) −
δq1 (s) · B · q100 (s) ds
= δq0 ·
∂q0
−∞
Next we impose initial conditions on (q1 , p1 )
q1 (s, T0 ) = q0 (T0 ),
p1 (s, T0 ) = 0
(s ≤ 0)
(4.12)
that is, the microscopic part of the motion is initially at rest. Then it can be seen
that the solution to the wave equation is given by (4.5) with φ1 (t) = q0 (t) and
φ2 = 0, that is
(
q1 (s + t − T0 , T0 ) = q0 (T0 ) if s + t ≤ T0
q1 (s, t) =
(4.13)
q0 (s + t)
if s + t ≥ T0
In particular, we see that q10 (0, t) = q̇1 (0, t) = q̇0 (t), and hence we obtain equations (4.1) and (4.2).
Thus it is seen that the irreversibility of equations of motion with damping comes
naturally from the special nature of the initial conditions. (And indeed any initial
condition satisfying q10 (s, T0 ) = q̇1 (s, T1 ) and q1 (0, T0 ) = q0 (T0 ) will work just as
well.)
5. The general equation of motion with damping
Suppose that H̃, and its corresponding Lagrangian L̃, have the same domains as
H0 and L0 . Here L̃ is also called the Rayleigh dissipation function.
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S. MONTGOMERY-SMITH
EJDE-2014/89
Consider the Hamiltonian on M = {(q0 , q1 ) ∈ M0 × M1 : q1 (0) = q0 } given by
Z 0
H(q, p) = H0 (q0 , p0 ) +
H̃(q1 (s), p1 (s)) + L̃(q1 (s), q10 (s)) ds
(5.1)
−∞
Note that the corresponding Lagrangian is
Z 0
L(q, q̇) = L0 (q0 , q̇0 ) +
L̃(q1 (s), q̇1 (s)) − L̃(q1 (s), q10 (s)) ds
(5.2)
−∞
Theorem 5.1. The equations of motion for the Hamiltonian given by (5.1) with
initial conditions (4.12) imply
q̇0 =
ṗ0 = −
∂H0
∂p0
∂H0
∂ L̃
(q0 , p0 ) −
(q0 , q̇0 )
∂q0
∂ q̇0
(5.3)
(5.4)
Proof. First, looking at the equations of motion for (q1 , p1 ) we obtain
ṗ1 = −
∂ H̃
∂ L̃
∂ ∂ L̃
0
(q1 , p1 ) −
(q1 , q10 ) +
(q
,
q
)
1
1
∂q1
∂q1
∂s ∂q10
q̇1 =
∂ H̃
(q1 , p1 )
∂p1
(5.5)
(5.6)
where the last term of equation (5.5) comes by integrating by parts, just as in the
previous section.
From the definition of L̃, we see that equation (5.6) is equivalent to
p1 =
∂ L̃
(q1 , q̇1 )
∂ q̇1
(5.7)
and substituting into equation (5.5) we obtain
∂ L̃
∂ L̃
∂ ∂ L̃
∂ ∂ L̃
(q1 , q10 ) −
(q1 , q̇1 ) −
(q1 , q̇1 ) =
(q1 , q10 )
0
∂t ∂ q̇1
∂q1
∂s ∂q1
∂q1
(5.8)
This is a wave equation, and since we have initial conditions (4.12), the solution (4.13) is valid. (The general solution is not so obvious: if L̃(x, y) = L̃(x, −y),
then φ1 (s + t) and φ2 (s − t) are both solutions to this equation. However it is
not obvious to me how these two solutions should be combined, since this wave
equation is non-linear.)
Next, the equations of motion for (q0 , p0 ) become
ṗ0 = −
∂H0
∂ L̃
(q1 (0), q10 (0))
(q0 , p0 ) − 0
∂q0
∂q1 (0)
∂H0
q̇0 =
(q0 , p0 )
∂p0
(5.9)
(5.10)
where the last term of equation (5.9) from the cross term from integrating by parts.
From equation (4.13) we see that q10 = q̇1 . Substituting this into equation (5.9),
and noting that q1 (0) = q0 , we obtain the desired result.
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HAMILTONIANS REPRESENTING EQUATIONS OF MOTION
7
Finally, it is worth noting that under the initial conditions (4.12) the Hamiltonian
and Lagrangian evaluate to
Z t
∂ L̃
H(q(t), p(t)) = H0 (q0 (t), p0 (t)) +
(q0 (τ ), q̇0 (τ )) · q̇0 (τ ) dτ
(5.11)
T0 ∂ q̇0
L(q(t), q̇(t)) = L0 (q0 (t), q̇0 (t))
(5.12)
In the case that L̃(x, y) = 12 y · B · y, the Hamiltonian becomes
Z t
q̇0 (τ ) · B · q̇0 (τ ) dτ
H(q(t), p(t)) = H0 (q0 (t), p0 (t)) +
(5.13)
T0
6. The arrow of time
Now that we have an explicit Hamiltonian that generates the equations of motion
with friction, we should examine whether it satisfies the comments made at the end
of Section 3 on the “arrow if time.”
One way to reverse the effect of friction would be to first solve the equation backwards in time, from t = T1 back to t = T0 , using initial conditions q1 (s, T1 ) = q0 (T1 ),
p1 (s, T1 ) = 0 for (s ≤ 0). Solve to find q1 (s, T0 ), noting that it will satisfy
q10 (s, T0 ) = −q̇1 (s, T0 ). Then this is the initial data that is required to make
the ‘wine glass come back together.’ Obviously not any initial data satisfying
q10 (s, T0 ) = −q̇1 (s, T0 ) will work, because the wave has to feed back into the equations of motion for q0 exactly the energy it needs to satisfy the reverse effect of
friction.
It would be interesting to see to what extent this equation is sensitive to the initial
data, perhaps illustrating that it is very hard to get it exactly right to get the ‘wine
glass to reconstruct.’ The idea would be to analyze the equations of motion for the
full system on M, but either restricting to solutions in which q̇1 = q10 (in which case
the solutions should be stable), or to the case in which q̇1 = −q10 (in which case
the solutions should be unstable — this latter case requires L̃(x, y) = L̃(x, −y),
otherwise the formulation of the problem would be harder). Almost surely one will
find that in the latter case, the equation is linearly unstable, corresponding to the
fact that the Hessian of L̃(x, ·) is positive definite.
7. Nöther’s Theorem
By applying Nöther’s Theorem to our Lagrangian, we can easily obtain a version
of Nöther’s Theorem that applies to dissipative systems. Suppose there is a flow
on M0 that preserves both L0 and L̃, let us denote the parameter driving the flow
by λ. Then the following quantity is conserved.
Z t
∂L1
∂q0 (t) ∂ L̃
∂q0 (τ ) π(t) =
(q0 (t), q̇0 (t)) ·
+
(q0 (τ ), q̇0 (τ )) ·
dτ (7.1)
∂ q̇0
∂λ λ=0
∂λ λ=0
T0 ∂ q̇0
Let us illustrate with a couple of examples of computing the momentum. Suppose
that q0 is n vectors in Rm : q0 = (q01 , q02 , . . . , q0n ). Let µx be the momentum in
the direction x
X
µ(t) =
mi q̇0i · x
(7.2)
i
which corresponds to the action
λ 7→ (q0 7→ (q0i + λx)ni=1 ))
(7.3)
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S. MONTGOMERY-SMITH
EJDE-2014/89
Suppose that
L0 (q0 , q̇0 ) =
X
2
1
2 mi |q̇0i |
+
XX
i
i
L̃(q0 , q̇0 ) = η
X
1
2 vij (|q0i
− q0j |)
(7.4)
j
mi 12 |q̇0i |2
(7.5)
i
where vij = vji . This gives the equation of motion
X (q0i − q0j )
0
mi q̈0i = −
vij
(|q0i − q0j |) − ηmi q̇0i
|q
−
q
|
0i
0j
j
(7.6)
Both of these Lagrangians are invariant under the action (7.3). We see that
Z t
π(t) = µx (t) + η
µx (τ ) dτ
(7.7)
T0
is conserved; that is,
µ̇x + ηµx = 0 ⇒ µx (t) = e−ηt µx (0)
Now suppose we change equation (7.5) to
∂
XX
1
L̃(q0 , q̇0 ) =
2 ηij ∂t |q0i − q0j |
i
j
(7.8)
(7.9)
where ηij = ηji , that is, all the damping effects are internal and depend completely
on how the distances between the particles change. The equation of motion is now
X (q0i − q0j )
X (q0i − q0j ) ∂
0
0
mi q̈0i = −
vij
(|q0i − q0j |) −
ηij
|q0i − q0j | (7.10)
|q0i − q0j |
|q0i − q0j |
∂t
j
j
This Lagrangian is also invariant under the action (7.3). We see that
X X (q0i − q0j ) ∂
∂ L̃
∂q0 (t) 0
=
ηij
|q0i − q0j | = 0
(q0 (t), q̇0 (t)) ·
∂ q̇0
∂λ λ=0
|q0i − q0j |
∂t
i
j
(7.11)
Therefore, π = µx , and thus the momentum µx is conserved.
Let us try a similar argument with angular momentum. Let αA be the angular
momentum about the anti-symmetric matrix A
X
α(t) =
mi q̇0i · A · q0i
(7.12)
i
which corresponds to the action
λ 7→ (q0 7→ (eλA · q0i )ni=1 ))
(7.13)
Since eλA is an orthogonal matrix, all of the above Lagrangians are invariant under
the action (7.3). For the case (7.5)
Z t
π(t) = αA (t) + η
αA (τ ) dτ
(7.14)
T0
is conserved; that is,
α̇A + ηαA = 0 ⇒ αA (t) = e−ηt αA (0)
(7.15)
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HAMILTONIANS REPRESENTING EQUATIONS OF MOTION
9
For the case (7.9), we obtain
X X (q0i − q0j ) ∂
∂ L̃
∂q0 (t) 0
=
ηij
|q0i − q0j | · A · q0i = 0
(q0 (t), q̇0 (t)) ·
∂ q̇0
∂λ λ=0
|q0i − q0j |
∂t
i
j
(7.16)
and thus similarly αA is conserved.
(We should add that technically Lagrangian (7.9) does not quite fit into our
framework, because it is not strictly convex in q̇0 . But this technicality is just
that, and easily overcome. One can either redo all the work, or add |q̇0 |2 to this
Lagrangian and take the limit as → 0.)
8. An identity
In this section we present an identity which we believe might be useful for creating
conserved quantities. Suppose that equation (5.8) is satisfied. Let
E = H̃(p1 , q1 ) − L̃(q1 , q10 )
F = q̇1 ·
∂ L̃
(q, q 0 )
∂q10
Then
∂F
∂E
=
(8.1)
∂t
∂s
In the case that L̃(x, y) = y · B(x) · y, so that L̃(q1 , q̇1 ) = H̃(q1 , p1 ), there is a
symmetry in the formulas so that we also have
∂E
∂F
=
∂s
∂t
and hence both E and F satisfy the standard wave equation:
∂2E
∂2E
=
,
∂t2
∂s2
∂2F
∂2F
=
∂t2
∂s2
(8.2)
(8.3)
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Stephen Montgomery-Smith
Department of Mathematics, University of Missouri, Columbia, MO 65211, USA
E-mail address: stephen@missouri.edu